Constructing the Complex Numbers

Question

This lecture on complex analysis constructs the complex numbers. At 55:08, the lecturer leaves the nature of the contradiction as an exercise.

I am trying to understand what the contradiction is. My understanding (after some thinking) is that by defintion: $$z=x+ye$$ And hence $$\sqrt{a+b^2/4}=e$$

and if this is real, then $z=x+ye$ itself must be completely real.

However, I am not sure. Can someone guide me what is the correct contradiction.

Note: I am quite new to Complex Analysis. Hence, this very basic question.

Answer 1

This video assumes $\Bbb R\cdot1\oplus\Bbb R\cdot e$ is a field and claims having "nearly proved" that this field must contain a square root for $-1.$ Its lengthy "proof" needs some cleaning to not look circular (and its "Case 1" was completely useless). You can replace it by the following (dropping the "$\cdot1$" and considering $\Bbb R$ as a subfield):

Let $a,b$ be the two real numbers s.t. $e^2=a+b\cdot e,$ and let $$f:=-\frac b2+e.$$ Then, $$f^2=a+\frac{b^2}4.$$ Therefore, $a+\frac{b^2}4<0,$ otherwise we would have $$\left(f-\sqrt{a+\frac{b^2}4}\right)\left(f+\sqrt{a+\frac{b^2}4}\right)=0$$ and derive $f=\pm\sqrt{a+\frac{b^2}4}\in\Bbb R,$ whence the following contradiction: $$e=\frac b2+f\in\Bbb R.$$

To conclude, simply check that $$\left(\frac f{\sqrt{\left|a+\frac{b^2}4\right|}}\right)^2=-1.$$

Answer 2

The value under the square root could be negative, and you'd then need to solve within the complex domain, i.e. it (the square root term as well as linear combinations of it like $x+ye)$ has complex roots.

However, if $a+\frac{b^2}{4}$ is positive, than it has a root $e=\sqrt{a+\frac{b^2}{4}}$ which is in the domain of the reals.

Anyway, it's a boring way to introduce the complex numbers. All you take out of it is: if the radicand - what's under the square root sign, is negative, then it - the radical - has complex roots.